1. Field of the Invention
The present invention relates to a MOS half-bridge drive circuit, particularly for power MOS half-bridges.
2. Discussion of the Related Art
Power MOS half-bridges generally include a pair of power MOS transistors, usually N-channel type, push-pull connected as shown by way of example in FIG. 1. FIG. 1 shows a half-bridge 1 including two MOS (typically VDMOS-Vertical Double Diffused MOS) transistors M.sub.1 and M.sub.2. Transistor M.sub.1 (top transistor) has its drain terminal connected to supply voltage VS, and its source terminal connected to the drain terminal of transistor M.sub.2 (bottom transistor), the source terminal of which is grounded. The gate terminals of transistors M.sub.1 and M.sub.2 are connected respectively to top and bottom drivers T and B which, in turn, are connected to a synchronizing input signal IN, for producing drive signals. One of transistors M.sub.1, M.sub.2 is turned off when the other is on.
The output of half-bridge 1 thus presents a signal VOU.sub.T, the amplitude of which switches between R.sub.DSon2 * I (where R.sub.DSon2 is the equivalent resistance between the drain and source of transistor M.sub.2 when saturated, and I is the output current) and V.sub.s -R.sub.DSon1 * I (where R.sub.DSon1 is the equivalent resistance between the drain and source of transistor M.sub.1 when saturated).
In half-bridge 1, to achieve a low voltage drop in whichever of transistors M.sub.1, M.sub.2 is on at the time, and therefore a high excursion of output voltage V.sub.OUT, the transistor must be driven by a high gate-source voltage (VGS), typically about 10 V, which provides for optimizing saturation resistance R.sub.DSon.
Such a high voltage is usually achieved without difficulty with respect to transistor M.sub.2, which permits the use of a reference voltage obtainable from the supply voltage. The situation is more complex, however, in the case of transistor M.sub.1, which requires a drive voltage of 10 V+V.sub.OUT, i.e. 10 V+V.sub.S -R.sub.DSon1 * I, which is higher than the supply voltage.
A highly effective conventional solution to the problem is to use a charge-pump or voltage doubling circuit, which gives a voltage higher than the supply voltage, and provides for driving one or more half-bridges. A simplified diagram of such a circuit, indicated by reference character 2, is shown in FIG. 2.
Charge-pep circuit 2 includes an oscillator 3; two condensers 4 and 5; and two diodes 6 and 7. Oscillator 3 is connected to a reference voltage V.sub.REF and ground, so that its output 8 switches continually between V.sub.REF and 0 V. In the first half-cycle, when output 8 is 0 V, condenser 4 charges to V.sub.S via diode 6; whereas, in the second half-cycle, when output 8 equals V.sub.REF, condenser 4 (whose terminal 9, not connected to oscillator 3, switches to voltage V.sub.REF +V.sub.S in relation to ground) transfers the charge to condenser 5. In the second half-cycle, diode 6 disconnects condensers 4 and 5 from the supply voltage V.sub.S, while, in the first half-cycle, in which condenser 4 restores the charge transferred to condenser 5, diode 7 disconnects condenser 5 from condenser 4 and the supply voltage V.sub.S.
Under normal operating conditions, if V.sub.CP is the output voltage of circuit 2 (drive voltage), and V.sub.be the voltage drop in diodes 6 and 7: EQU V.sub.CP =V.sub.S -2 i V.sub.be +V.sub.REF ( 1)
and, assuming in this particular case that V.sub.be =0.7V and V.sub.REF =10 V: EQU V.sub.CP =V.sub.S -1.4V+10 V.apprxeq.V.sub.S +9.6V
Charge-pump circuit 2 therefore provides a drive voltage ensuring a sufficient voltage drop V.sub.GS in transistor M.sub.1 to effectively saturate it.
Reference voltage V.sub.REF is generally produced using circuit 10 in FIG. 3, which presents a minimum voltage difference D.sub.V between supply and reference voltages V.sub.S and V.sub.REF. Circuit 10 includes a bipolar NPN transistor 11; a resistor 12 connected between the base and collector of transistor 11; and two diodes and 14, of which diode 13 is a Zener diode, connected in series between the base of transistor 11 and ground. The emitter of transistor 11 is connected to supply voltage V.sub.S, and the base defines the output of circuit 10.
If R, in circuit 10, is the resistance of resistor 12; h.sub.fe the current gain for small signals of transistor 11; and I.sub.O the emitter current of transistor 11; when, as in this case, V.sub.S &lt;V.sub.Z +V.sub.be (where V.sub.Z and V.sub.be are the voltage drop in Zener diode 13 and diode 14 respectively): EQU D.sub.V =V.sub.S -V.sub.REF =(R*I.sub.O)/h.sub.fe +V.sub.be
which, by appropriate sizing of the components, may be reduced to 1 V, so that: EQU V.sub.REF =V.sub.S -D.sub.V =1V (2)
There are some situations, however, in which the circuit shown in FIG. 3 is impractical, as in the case of automotive applications, which require that the MOS half-bridge, operating normally with a supply voltage V.sub.S of 6 V, should also be capable of operating with a minimum supply voltage V.sub.Smin of 5.4 V, i.e. that the drive circuit of top transistor M.sub.1 supply an output voltage V.sub.CP of at least roughly 9.3 V.
In fact, to achieve a fairly good saturation resistance, V.sub.DMOS transistors must be driven in such a manner as to present a voltage drop V.sub.GS of at least 4 V. This is obtained easily, in the case of transistor M.sub.2, even with a minimum supply voltage V.sub.Smin. The same does not apply, however, to transistor M.sub.1, which, with a minimum supply voltage V.sub.Smin of 5.4 V, requires that V.sub.OUT .apprxeq. 5.3 V and, therefore, V.sub.CP .apprxeq. 9.3 V.
According to equation (2), however, the circuit shown in FIG. 3 circuit provides, at most, supplying an output voltage V.sub.REF of 4.4 V, so that, according to equation (1), charge-pump circuit 2 provides solely for supplying a voltage V.sub.CP of: EQU V.sub.CP =5.4 V-1.4 V+4.4 V=8.4 V
The source-drain voltage drop of transistor M.sub.1 therefore equals: EQU V.sub.GS1 =V.sub.CP -V.sub.OUT =3.1 V
which is well below the required value.
Another known solution for generating the reference voltage V.sub.REF includes using a PNP transistor in place of transistor 11, as shown in FIG. 4, where PNP transistor 15 has an emitter connected to supply voltage V.sub.S, and a collector defining the output of the circuit. A resistor 16 is connected parallel to the base-emitter junction of transistor 15; a current source 17 is connected between the base of transistor 15 and ground; and a Zener diode 18 is connected between the output and ground.
The above solution, however, despite providing for a voltage drop D.sub.V =V.sub.S -V.sub.REF equal to the emitter-collector saturation voltage of transistor 15 (and, therefore, below that of the circuit shown in FIG. 3), is not used for the following reasons:
a) the maximum output current I.sub.OUT required by the load must be supplied at all times by transistor 15, even when not needed (thus resulting in high power consumption and dissipation); and
b) for a given area, a PNP transistor provides for carrying a much smaller current than an NPN type, and the Zener diode must be sized to withstand all the current (thus increasing the size of the integrated circuit).
It is, therefore, an object of the present invention to provide a MOS half-bridge drive circuit enabling the half-bridges to operate even with a low supply voltage.